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प्रश्न
Find the values of k for which the given quadratic equation has real and distinct roots:
kx2 + 6x + 1 = 0
उत्तर
The given quadric equation is kx2 + 6x + 1 = 0, and roots are real and distinct.
Then find the value of k.
Here,
a = k, b = 6 and c = 1
As we know that D = b2 - 4ac
Putting the value of a = k, b = 6 and c = 1
D = (6)2 - 4 x (k) x (1)
= 36 - 4k
The given equation will have real and distinct roots, if D > 0
36 - 4k > 0
Now factorizing of the above equation
36 - 4k > 0
4k < 36
k < 36/4
k < 9
Now according to question, the value of k less than 9
Therefore, the value of k < 9.
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Solution :
Compare x2 + 2x – 9 = 0 with ax2 + bx + c = 0
a = 1, b = 2, c = `square`
∴ b2 – 4ac = (2)2 – 4 × `square` × `square`
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∴ The roots of the equation are real and unequal.
